What is the derivative of f(x)=e^{1-2x}(x+cos(2x)) ?

The derivative of f(x)=e^{1-2x}(x+cos(2x)) is -2e^{1-2x}(x+cos(2x))+e^{1-2x}(1-2sin(2x))

What is the first derivative of f(x)=e^{1-2x}(x+cos(2x)) ?

The first derivative of f(x)=e^{1-2x}(x+cos(2x)) is -2e^{1-2x}(x+cos(2x))+e^{1-2x}(1-2sin(2x))

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derivative of

f (x) = e^{1 - 2 x} (x + cos (2 x))

- 2 e^{1 - 2 x} (x + cos (2 x)) + e^{1 - 2 x} (1 - 2 sin (2 x))

Solution steps

\frac{d}{d x} (e^{1 - 2 x} (x + cos (2 x)))

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = e^{1 - 2 x}, g = x + cos (2 x)

= \frac{d}{d x} (e^{1 - 2 x}) (x + cos (2 x)) + \frac{d}{d x} (x + cos (2 x)) e^{1 - 2 x}

\frac{d}{d x} (e^{1 - 2 x}) = - 2 e^{1 - 2 x}

\frac{d}{d x} (x + cos (2 x)) = 1 - sin (2 x) \cdot 2

= (- 2 e^{1 - 2 x}) (x + cos (2 x)) + (1 - sin (2 x) \cdot 2) e^{1 - 2 x}

Simplify

= - 2 e^{1 - 2 x} (x + cos (2 x)) + e^{1 - 2 x} (1 - 2 sin (2 x))

f (x) = x^{3} + x^{2} + 1, g (x) = 2 x + 1

\int \frac{1}{25 - 36 x ^{2}} d x

n = 4 \sum \infty e^{5 - 6 n}

\frac{\partial}{\partial y} (y (1 + ln (x y)))

t y^{^{'}} + t y = 1 - y, y (1) = 0

What is the derivative of f(x)=e^{1-2x}(x+cos(2x)) ?
The derivative of f(x)=e^{1-2x}(x+cos(2x)) is -2e^{1-2x}(x+cos(2x))+e^{1-2x}(1-2sin(2x))
What is the first derivative of f(x)=e^{1-2x}(x+cos(2x)) ?
The first derivative of f(x)=e^{1-2x}(x+cos(2x)) is -2e^{1-2x}(x+cos(2x))+e^{1-2x}(1-2sin(2x))